| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Singular matrix conditions |
| Difficulty | Moderate -0.8 This is a straightforward FP1 question testing basic matrix concepts: finding when det(X)=0 (routine calculation), computing a 2×2 inverse using the standard formula, and applying the inverse transformation. All parts are direct applications of standard procedures with no problem-solving insight required, making it easier than average but not trivial since it involves Further Maths content. |
| Spec | 4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03l Singular/non-singular matrices4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Determinant: \(2 - 3a = 0\) and solve for \(a =\) | M1 | |
| So \(a = \frac{2}{3}\) or equivalent | A1 | (2) |
| (b) Determinant: \((1 \times 2) - (3 \times -1) = 5\) | M1A1 | \((\Delta)\) |
| \(\mathbf{Y}^{-1} = \frac{1}{5}\begin{pmatrix} 2 & 1 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} 0.4 & 0.2 \\ -0.6 & 0.2 \end{pmatrix}\) | (2) | |
| (c) \(\frac{1}{5}\begin{pmatrix} 2 & 1 \\ -3 & 1 \end{pmatrix}\begin{pmatrix} 1 & -\lambda \\ 7\lambda - 2 \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 2 - 2\lambda + 7\lambda - 2 \\ -3 + 3\lambda + 7\lambda - 2 \end{pmatrix} = \begin{pmatrix} \lambda \\ 2\lambda - 1 \end{pmatrix}\) | M1depM1A1 A1 | (4) [8] |
| Answer | Marks |
|---|---|
| \(\begin{pmatrix} 1 & -1 \\ 3 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 - \lambda \\ 7\lambda - 2 \end{pmatrix}\) so \(x - y = 1 - \lambda\) and \(3x + 2y = 7\lambda - 2\) | M1M1 |
| Solve to give \(x = \lambda\) and \(y = 2\lambda - 1\) | A1A1 |
(a) Determinant: $2 - 3a = 0$ and solve for $a =$ | M1 |
So $a = \frac{2}{3}$ or equivalent | A1 | (2)
(b) Determinant: $(1 \times 2) - (3 \times -1) = 5$ | M1A1 | $(\Delta)$
$\mathbf{Y}^{-1} = \frac{1}{5}\begin{pmatrix} 2 & 1 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} 0.4 & 0.2 \\ -0.6 & 0.2 \end{pmatrix}$ | | (2)
(c) $\frac{1}{5}\begin{pmatrix} 2 & 1 \\ -3 & 1 \end{pmatrix}\begin{pmatrix} 1 & -\lambda \\ 7\lambda - 2 \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 2 - 2\lambda + 7\lambda - 2 \\ -3 + 3\lambda + 7\lambda - 2 \end{pmatrix} = \begin{pmatrix} \lambda \\ 2\lambda - 1 \end{pmatrix}$ | M1depM1A1 A1 | (4) [8]
**Alternative method for (c):**
$\begin{pmatrix} 1 & -1 \\ 3 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 - \lambda \\ 7\lambda - 2 \end{pmatrix}$ so $x - y = 1 - \lambda$ and $3x + 2y = 7\lambda - 2$ | M1M1 |
Solve to give $x = \lambda$ and $y = 2\lambda - 1$ | A1A1 |
**Notes:**
(b) M for $\frac{1}{\text{their det}}\begin{pmatrix} 2 & 1 \\ -3 & 1 \end{pmatrix}$
(c) First M for their $\mathbf{Y}^{-1}\mathbf{B}$ in correct order with B written as a 2×1 matrix; second M dependent on first for attempt at multiplying their matrices resulting in a 2×1 matrix; first A for $\lambda$; second A for $2\lambda - 1$
**Alternative for (c):**
First M to obtain two linear equations in $x, y, \lambda$; Second M for attempting to solve for $x$ or $y$ in terms of $\lambda$
---6. $\mathbf { X } = \left( \begin{array} { l l } 1 & a \\ 3 & 2 \end{array} \right)$, where $a$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$ for which the matrix $\mathbf { X }$ is singular.
$$\mathbf { Y } = \left( \begin{array} { r r }
1 & - 1 \\
3 & 2
\end{array} \right)$$
\item Find $\mathbf { Y } ^ { - 1 }$.
The transformation represented by $\mathbf { Y }$ maps the point $A$ onto the point $B$.\\
Given that $B$ has coordinates ( $1 - \lambda , 7 \lambda - 2$ ), where $\lambda$ is a constant,
\item find, in terms of $\lambda$, the coordinates of point $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q6 [8]}}